I'm trying to compute the total Chern class of the tangent bundle of a manifold constructed as a "compactification" or "projective completion", but I'm not sure I quite have all the details under control.
The manifold $M_{8}$ is eight-dimensional and complex, and constructed as a two-sphere bundle over a six-dimensional (Kahler) base space $B_{6}$. I want to think of this bundle as the total space of the anti-canonical line bundle $\mathcal{L}$ of $B_{6}$, together with a point added at infinity in each $\mathbb{C}$ fibre to make them into spheres. In practice, from here, I believe this means I can think of $M_{8}$ as the total space of $\mathbb{P}(\mathcal{O}\oplus\mathcal{L})$, where $\mathbb{P}$ indicates the projectivisation and $\mathcal{O}$ is the trivial line bundle over $B_{6}$.
I then want to compute the total Chern class of $M_{8}$ in terms of the data of $B_{6}$. As a start, following this, I have that, taking $V=\mathcal{O}\oplus\mathcal{L}$ and $\pi\colon\mathbb{P} V\to B_{6}$, $$ c(T_{\mathcal{M}_{8}})=c(\pi^{*}(\mathcal{O}\oplus\mathcal{L})\otimes\gamma^{*})c(\pi^{*}T_{B_{6}}), $$ where $\gamma$ is the tautological line bundle over $\mathbb{P} V=\mathcal{M}_{8}$. Since $\mathcal{L}$ is the anti-canonical bundle of $B_{6}$, I also know that $c_{1}(\mathcal{L})=c_{1}(T_{B_{6}})$. The Whitney sum formula, etc., then gives \begin{align*} c(T_{\mathcal{M}_{8}}) & =c(\pi^{*}\mathcal{O}\otimes\gamma^{*})c(\pi^{*}\mathcal{L}\otimes\gamma^{*})c(\pi^{*}T_{B_{6}})\\ & =\left(1+\pi^{*}c_{1}(\mathcal{O})+c_{1}(\gamma^{*})\right)\left(1+\pi^{*}c_{1}(T_{B_{6}})+c_{1}(\gamma^{*})\right)\pi^{*}c(T_{B_{6}}), \end{align*} which I can expand out to read off the Chern classes. At this point, my geometry runs out. I have two questions:
- Naively, I'd say that $\gamma^{*}=\pi^{*}\mathcal{O}(1)$ and would then need to know $\mathcal{L}$ in terms of $\mathcal{O}(n)$ in order to make progress. Does that sound correct, or is that too quick? From this, it seems that it might be more complicated...
- In fact, the end goal is to compute the Pontryagin classes for $\mathcal{M}_{8}$. In particular, is it obvious that $p_{1}^{2}$ and $p_{2}$ are always trivial, since $\mathcal{M}_{8}$ is constructed as a bundle? (Since everything reduces to classes $B_{6}$, there are no non-trivial eight-form classes constructed from the Chern classes?)
Thank you in advance, and sorry if I've said something imprecise. I think this is fairly standard material, but I haven't come across it before.
Edit: I think the point is to use the tautological exact sequence $$ 0 \to \gamma \to \pi^* V \to T_{\mathbb{P}(V)/B_6}\otimes \gamma \to 0 $$ and then the observation that the Chern classes of $V$ satisfy (Bott & Tu, page 270) $$ x^n + \pi^* c_1(V) x^{n-1} + \dots + \pi^* c_n(V) =0 $$ where $x = c_1(\gamma^*)$ and $V$ is rank $n$. For $V=\mathcal{O} \oplus \mathcal{L}$, we then have $$ x^2 + \pi^* c_1(\mathcal{L})x=0 $$ so that $x = c_1(\gamma^*) = - \pi^* c_1(\mathcal{L})$. (In particular, $\gamma^* \neq \pi^*\mathcal{O}(1)$, but instead picks up a dependence on $\mathcal{L}$.) With this in hand, one can use the above formula for $c(T_{\mathcal{M}_{8}})$ to compute the Chern classes.